62.Unique_Paths_(Medium).md

62. Unique Paths (Medium)

Date and Time: Jun 26, 2024, 16:45 (EST)

Link: https://leetcode.com/problems/unique-paths/

Question:

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to $2 * 10^9$.

Example 1:

Input: m = 3, n = 7

Output: 28

Example 2:

Input: m = 3, n = 2

Output: 3

Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:

1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

KeyPoints:

We use DP to store how many ways for each grid to reach the bottom-right corner. For the bottom row and the right most column, they only have one way to reach the goal, so it is one for them.

Python DP Optimized Solution:

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        prev = [1] * n
        for r in range(m-2, -1, -1):
            dp = [1] * n
            for c in range(n-2, -1, -1):
                dp[c] = prev[c] + dp[c+1]
            prev = dp
        return prev[0]

Time Complexity: $O(m * n)$
Space Complexity: $O(n)$, we only store the previous row.

Python DP Solution:

Same idea, but we just build the dp table first

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[1] * n for _ in range(m)]
        for r in range(m-2, -1, -1):
            for c in range(n-2, -1, -1):
                dp[r][c] = dp[r+1][c] + dp[r][c+1]
        return dp[0][0]

Time Complexity: $O(m * n)$
Space Complexity: $O(m * n)$, because we store the whole m * n dp table.